What does $\mathfrak{m}(\varprojlim R/\mathfrak{m}^i)$ mean?

Question

Let $\mathfrak{m}$ be an ideal of the ring $R$, and consider it as an $R$-module. What does $\mathfrak{m}(\varprojlim R/\mathfrak{m}^i)$ mean? In the limit the morphisms are the natural projections.

I know about the explicit description of $\varprojlim R/\mathfrak{m}^i$, a general element looks like:

$$(0+R,c_1+\mathfrak{m},c_2+\mathfrak{m}^2,\cdots)$$ where $\pi(c_i+\mathfrak{m}^i)=c_{i-1}+\mathfrak{m}^{i-1}$. However, I don't see what the $\mathfrak{m}$ in front the limit does?

Are these just elements of the form

$$m(0+R,c_1+\mathfrak{m},c_2+\mathfrak{m}^2,\cdots)=(0+R,mc_1+\mathfrak{m},mc_2+\mathfrak{m}^2,\cdots)=(0+R,0+\mathfrak{m},mc_2+\mathfrak{m}^2,\cdots)$$ Or maybe they are finite combinations of such elements? Thanks

Answer 1

There exists a natural map from $R$ to the inverse limit $\varprojlim (R/\mathfrak{m}^i)$ that takes every element $r$ in $R$ to the sequence with all entries equal $r$ . Now, $m$ is an ideal of $R$ so $m \varprojlim (R/\mathfrak{m}^i)$ should be the extension of this ideal from $R$ to $ \varprojlim (R/\mathfrak{m}^i)$. So, by definition, it should be combination of those elements ( sums). It is contained in the ideal of $ \varprojlim (R/\mathfrak{m}^i)$ consisting of all sequences whose elements are in $m$ at each component. In some cases they do equal.

Answer 2

Given an ideal $I\subseteq R$ and an $R$-module $M$, $IM$ is defined as the submodule of $M$ generated by products $im$ where $i\in I$ and $m\in M$. Explicitly, $IM$ is the set of elements of $M$ which can be writte as a sum $\sum_{k=1}^n i_k m_k$ for some $n\in\mathbb{N}$, some $i_1,\dots,i_n\in I$ and some $m_1,\dots, m_n\in M$. In this case, $I=\mathfrak{m}$ and $M=\varprojlim R/\mathfrak{m}^i$ (with the $R$-module structure on $M$ given coordinatewise). Note that $\varprojlim R/\mathfrak{m}^i$ can also be given the structure of a ring (with coordinatewise multiplication), and you can also describe this as the ideal in $\varprojlim R/\mathfrak{m}^i$ generated by the image of $\mathfrak{m}$ under the homomorphism $R\to \varprojlim R/\mathfrak{m}^i$ taking an element $r\in R$ to $(r+R,r+\mathfrak{m},r+\mathfrak{m}^2,\dots)$.